Question

in a single isolated atom of hydrogen, electrons make transition from 4th excited state to ground stateproducing maximum possible number of wavelengths. If the 2nd lowest energy photon is used to furtherexcite an already excited sample of Li2+ ion, then transition will be:(A) 12 -15(B) 9-12(C) 6-9(D) 3-6​

Answer 1

Answer:

From 9 to 12.

Explanation:

Since, we know that the single electronic species are those which can be solved by using Rydberg's equation and both hydrogen and lithium are single electronic species hence using Rydberg's equation we can solve the question.

So, we know that the energy levels of the bohr and the hydrogen can be determined by using the formulae as :-

E = -r*z^2/n^2 .

E=-r/(4)^2=-9r/n^2.

Hence on solving this equation we will get the value of the n=12. So, the answer will be for lithium the value will be from n=9 to n=12.